1969 AMC 12 第 28 题

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28.

在半径为 11 的圆内部,设满足下列条件的点 PP 的个数为 nnPP 到某一直径两个端点的距离平方和为 33。则 nn 为:

Let nn be the number of points PP interior to the region bounded by a circle with radius 1,1, such that the sum of the squares of the distances from PP to the endpoints of a given diameter is 3.3. Then nn is:

00

11

22

44

无穷多个

infinite

答案:E
知识点:坐标几何距离公式
难度评级:1720
小提示:

将圆心置于原点,并取直径端点为 (1,0)(-1,0)(1,0)(1,0)

Place the circle at the origin with diameter endpoints (1,0)(-1,0) and (1,0)(1,0)

大提示:

化简两个距离的平方和

Simplify the sum of the two squared distances

解答:

P=(x,y)P=(x,y),并取直径端点为 (1,0)(-1,0)(1,0)(1,0)。条件化为 PA2+PB2=3,2x2+2y2+2=3 \begin{aligned} PA^2+PB^2&=3,\\ 2x^2+2y^2+2&=3 \end{aligned}\text{。}因此 x2+y2=12x^2+y^2=\frac{1}{2}。这是一整个位于给定单位圆内部、半径为 12\frac{1}{\sqrt2} 的圆,包含无穷多个点。

所以正确答案是 E

Let P=(x,y)P=(x,y) and take the diameter endpoints as (1,0)(-1,0) and (1,0).(1,0). The condition becomes PA2+PB2=3,2x2+2y2+2=3. \begin{aligned} PA^2+PB^2&=3,\\ 2x^2+2y^2+2&=3. \end{aligned} Thus x2+y2=12.x^2+y^2=\frac{1}{2}. This is an entire circle of radius 12,\frac{1}{\sqrt2}, lying inside the given unit circle. It contains infinitely many points.

Therefore, the correct answer is E.

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